Q.Standard molar enthalpy of formation, ΔfH° is just a special case of enthalpy of reaction, ΔrH°. Is the ΔrH° for the following reaction same as ΔfH°? Give reason for your answer.
CaO(s) + CO2(g) → CaCO3(s); ΔfH° = -178.3 kJ mol^-1

The standard molar enthalpy of formation ($\Delta_f H^\circ$) is a special case of enthalpy of reaction.

Concept: Standard Molar Enthalpy of Formation ($\Delta_f H^\circ$)

1. The standard molar enthalpy of formation ($\Delta_f H^\circ$) is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their most stable standard states.
2. For the given reaction, $\text{CaO}(s) + \text{CO}_2(g) \rightarrow \text{CaCO}_3(s)$, one mole of $\text{CaCO}_3(s)$ is formed.
3. However, the reactants, $\text{CaO}(s)$ and $\text{CO}_2(g)$, are compounds, not the constituent elements of $\text{CaCO}_3(s)$ (which are $\text{Ca}(s)$, $\text{C}(s, \text{graphite})$, and $\text{O}_2(g)$) in their standard states. …

In thermochemistry, it's crucial to distinguish between different types of enthalpy changes. While the standard molar enthalpy of formation ($\Delta_f H^\circ$) is indeed a special case of the standard enthalpy of reaction ($\Delta_r H^\circ$), not every reaction enthalpy qualifies as a formation enthalpy. The distinction lies in the specific conditions defining the formation reaction.

Let's break down the concepts and then apply them to the given reaction.

Understanding Enthalpy Changes

1. Standard Enthalpy of Reaction ($\Delta_r H^\circ$):

   This is the enthalpy change when a reaction occurs under standard conditions (usually $298.15 \text{ K}$ and $1 \text{ bar}$ pressure, with all substances in their standard states). It represents the heat absorbed or released during a chemical reaction.

2. Standard Molar Enthalpy of Formation ($\Delta_f H^\circ$):

   This is a very specific type of standard enthalpy of reaction. It is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their most stable physical states (standard states) under standard conditions.

   For example, the standard state of oxygen is $\text{O}_2(g)$, carbon is $\text{C}(graphite)$, and calcium is $\text{Ca}(s)$.

   Important

   For a reaction to represent the standard molar enthalpy of formation of a compound, two conditions must be met:

   1. Exactly one mole of the compound must be formed.
   2. The reactants must be the constituent elements of that compound, each in its standard state.

Step-by-Step Analysis

1. Analyze the given reaction: The reaction provided is:

The value given, $\Delta_f H^\circ = -178.3 \text{ kJ mol}^{-1}$, is actually the standard enthalpy of reaction ($\Delta_r H^\circ$) for this specific process. The notation $\Delta_f H^\circ$ here is misleading if it implies formation enthalpy of $\text{CaCO}_3(s)$. It simply states the enthalpy change for the reaction is $-178.3 \text{ kJ mol}^{-1}$.

2. Determine the standard molar enthalpy of formation ($\Delta_f H^\circ$) for $\text{CaCO}_3(s)$:

To find the standard molar enthalpy of formation of $\text{CaCO}_3(s)$, we must write the reaction where one mole of $\text{CaCO}_3(s)$ is formed from its constituent elements in their standard states.

* The elements in $\text{CaCO}_3$ are Calcium ($\text{Ca}$), Carbon ($\text{C}$), and Oxygen ($\text{O}$).

* Their standard states are: $\text{Ca}(s)$, $\text{C}(graphite)$, and $\text{O}_2(g)$.

Therefore, the formation reaction for $\text{CaCO}_3(s)$ is:

The enthalpy change for this specific reaction is defined as $\Delta_f H^\circ (\text{CaCO}_3, s)$.

3. Compare the two reactions:

Let's compare the reactants in the given reaction with those in the formation reaction:

| Feature                 | Given Reaction: $\text{CaO}(s) + \text{CO}_2(g) \rightarrow \text{CaCO}_3(s)$ | Formation Reaction: $\text{Ca}(s) + \text{C}(graphite) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{CaCO}_3(s)$ |
| :---------------------- | :----------------------------------------------------------------------- | :---------------------------------------------------------------------------------------------------- | …